Examining whether the relation "$aRb$ iff $a + 2b \equiv 0 \pmod 3$" is reflexive, symmetric, antisymmetric, or transitive [duplicate]

Question

Have I shown correctly which properties the relation fulfills?

$$aRb \text{ iff } a + 2b \equiv 0 \pmod 3$$

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$(1)$ Reflexivity

Set $b=a$

$a + 2a = 3a \equiv 0 \pmod 3$

Hence, the relation is reflexive.

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$(2)$ Symmetry

(*) $a + 2b \equiv 0 \pmod 3$

(**) $b + 2a \equiv 0 \pmod 3$

Assuming (*) and (**) are true, the sum must also be divisible by three (if * and ** were not satisfied, the sum would lie in another congruence class):

$a + b + 2a + 2b = (a+b) + 2(a+b) \equiv 0 \pmod 3$

Set $a+b=c$

Then $c + 2c = 3c \equiv 0 \pmod 3$

Hence, the relation is symmetric.

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$(3)$ Antisymmetry

Counterexample: $a=4, b = -5$

$4 + 2*(-5)=-6 \equiv 0 \pmod 3\implies aRb$

$-5 + 2*4 = 3 \equiv 0 \pmod 3\implies bRa$

But $a\neq b$, hence the relation is not antisymmetric.

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$(4)$ Transitivity

$a + 2b \equiv 0 \pmod 3$

$b + 2c \equiv 0 \pmod 3$

$a + b + 2(b+c) \equiv 0 \pmod 3$

$(a + 2b) + b + 2 c \equiv 0 \pmod 3$

Hence, $(a+2c)\equiv0 \pmod 3 \implies $ relation is transitive.

Answer 1

The symmetry proof is nonsense. You're assuming that $aRb$ and $bRa$ to be true at the outset, and reach an unclear conclusion.

The definition of symmetry is: if $aRb$ is true, then $bRa$ is also true.

For example:

• $xRy$ for $x,y \in \mathbb{R}$ is true when $|x-y|=1$
• $xRy$ for $x,y \in \mathbb{R}^n$ is true when $\langle x,y \rangle = 0$
• $xRy$ for $x,y$ lines in $\mathbb{R}^2$ is true when $x \parallel y$

These are symmetric relations. For instance, $$ 1 = |x-y| = |(-1)(y-x)| = |-1| \cdot |y-x| = |y-x| $$ Thus, you start out by only assuming that $aRb$ holds, and from that alone you need to show that $bRa$ holds.

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For the transitivity proof, you reach the right conclusion, but I don't think you understand what you're doing based on the equations written.

So, we know that $aRb$ and $bRc$ are true, and we seek to prove that $aRc$ is true. Since $aRb$ and $bRc$, then $$\begin{align*} a + 2b &\equiv 0 \pmod 3 \\ b + 2c &\equiv 0 \pmod 3 \end{align*}$$ The natural way forward, to proving that $$ a + 2c \equiv 0 \pmod 3 $$ is to add the two equations we know to be true. This gives us $$ a+2b+b+2c \equiv 0 \pmod 3 $$ But this is the same as $$ a+2c+3b \equiv 0 \pmod 3 $$ Since, presumably, $a,b,c \in \mathbb{Z}$, then $3b \equiv 0 \pmod 3$ always, leaving us with $$ a+2c \equiv 0 \pmod 3 $$ and hence $aRc$ as desired.

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The other two items are fine and I see no prominent issues with them.